Cantor Confusion
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From: mueckenh on 20 Oct 2006 16:53 jpalecek(a)web.de schrieb: > mueckenh(a)rz.fh-augsburg.de napsal: > > jpalecek(a)web.de schrieb: > > > > > > > > > > > > 0. > > > > > > > > /a \ > > > > > > > > 0 1 > > > > > > > > /b \c / \ > > > > > > > > 0 1 0 1 > > > > > > > > ..................... > > > > > > > > > > > > > > An edge is related to a set of path. If the paths, belonging to this > > > > set, split in two different subsets, then the edge related to the > > > > complete set is divided and half of that edge is related to each of the > > > > two subsets. If it were important, which parts of the edges were > > > > related, then we could denote this by "edge a splits into a_1 and a_2". > > > > But because it is completely irrelevant which part of an edge is > > > > related to which subset, we need not denote the fractions of the edges. > > > > > > Sorry, but your "proof" doesn't work. Imagine an infinite path in the > > > tree. Which is the edge it inherits as a whole? Whenever you give me > > > that edge, I can tell you're lying because if a path inherits an edge > > > as a whole, it means that the path terminates by that edge. > > > > How should I be able to name the last term of a sum which has no last > > Indeed, INDEED. But that means, that there is no term 1 in the series, > too. No. That does it not mean. Unless you agree that no real numbers do exist. > Moreover, it means that there is no positive term in the series. > (Because > 1 would be counted for the last edge, which does not exist, etc.). > > > term? But while we cannot name any individual edge we can prove: No > > path splits into two paths without the supply of two new edges, one > > edge for each path. This implies there cannot be less edges than paths. > > No. This holds for the finite case only. Why? If you follow a path you can see that where ever it splits off another path, this splitting happens by an edge. As this situation does never change, it remains in infinity for the whole path. For every partial sequence (i.e. for every finite sequence of the path with n edges, starting from the top), we have the "load of edges" accumulated: (1 - (1/2)^n)/(1 - 1/2). We cannot construct a finite sequence with less edges, even if we would like to do so. In the limit n --> oo we obtain the whole load 2. This is the correct calculation. All other arguing is nonsense, with or without marriage. > > > Or the other way round: Assume there were more paths than edges, then > > at least two paths could no be distinguished. (A path can be > > distinguished from every other path by at least one edge.) > > Which doesn't mean anything. Oh yes, a different edge is obviously a necessary condition to distinguish two paths. > See the marriage theorem (or Hall's > theorem) for the conditions you would have to fulfill to have an > injective > mapping of paths to their edges). And note, you cannot use marriage > theorem to prove existence of any injective mapping of paths to edges). > > > > This is > > > impossible for infinite paths. > > > > Of course that is impossible. Therefore the sum 1 + 1/2 + 1/4 + ... is > > This series is irrelevant. This kind of arguments accumulate here. > > > a infinite sum. But nevertheless your argument covers only half of the > > story. Whenever you give me two infinite paths, I can name an edge > > which belongs to only one of them. > > Again, this doesn't buy you anything. See marriage theorem. This kind of arguments accumulate here. > > > > The same argument applies to other terms > > > in the sum. (That edge is inherited by an infinite path by 1/1024! > > > Ok, but that means that the path terminates 10 levels lower). This > > > means that infinite path inherit zero edges in your proof. > > > > Then the series 1 + 1/2 + 1/4 + ... contains zeros? > > No, that means that series is a wrong one for your problem. The correct > one is 0+0+0+... The correct answer is then: there are no paths there are no real numbers. > > > The distance between any two edges of one path is infinite? > > The distance of any edge from the end of an infinite path is infinite. Therefore we have an infinite series with value 2. Regards, WM
From: MoeBlee on 20 Oct 2006 16:58 mueck...(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > But the understanding of them is a prerequisite of understanding > > > infinite sets (according to Cantor who new more about that matter en > > > Zermelo, Fraenkel and v. Neuman together. Therefore it is no wonder > > > that set theorists easily intermingle both). > > > > Understanding of them is a philosophical concern, and such > > philosophical inquiries are not precluded by set theory; only that the > > philosophical investigations of differences between actual and > > potential infinity do not occur WITHIN set theory; > > This shows that set theory is completely useless, because the only > topic to be dealt with successfully is infinity. I don't get you. Set theory gives an axiomatization for analysis. That's useful. That you don't want to use it is fine. But it doesn't make it useless. > > Okay, so I take it that your argument about that binary tree is meant > > to be within say, Z set theory. Yes, I do wish to be educated by you. > > Rather than just giving an ostensive definition of the relation with > > edges and paths "continuing in this manner to infinity", would you > > please give a rigorous definition of the relation? > > "Continue in this manner" is just what is used to describe how the > infinite is realized. > > Consider a finite path with n edges. This path carries a load of (1 - > (1/2)^n)/(1 - 1/2) own edges. For n --> oo this yields a load of 2 > edges. It is very simple, it derived from current mathematics, which is > derived from your axioms, therefore it must not further be confused by > retranslating it with your axioms. Okay, now that I asked for a definition of the relation you mentioned, you're not giving that definition, but instead giving a combinatorical/numerical argument with more terminology. What is a "load of edges"? What is the definition of "a path carries a load of edges"? If this is standard terminology in graph theory, then please forgive my ignorance and supply me with the standard definition. If it is not standard terminology, then please give me your own definition. > Please follow my discussion with jpale who understood the problem. I'd like to be specific as to these definitions without complications from another conversation. But if you don't want to supply the definitions and explanations in our exchanges, then fine, but then I'll not be in a position to comment on your arguments. > > The diagonal argument does not contradict that 1=.999... > > No, it is even unaware of this fact, because the necessary convergence > is missing. You don't know what your're talking about. The convergence is proven. MoeBlee
From: mueckenh on 20 Oct 2006 17:00 Dik T. Winter schrieb: > In article <1161276322.150252.120060(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > Again replying to more than one article at once without giving proper > > > references. That makes it very difficult to follow threads. > > > > I am sorry, I have only 15 shots per several hours, but far more opponents. > > Why do you think that is the case? I have not the slightest idea. > > > I repeat only that part of the discussion which I answer to. > > Yes, and meantime you are breaking threads so that it becomes difficult to > see what you are actually replyint to. > > > > Yes, according to Cantor. And according to modern set theory that is > > > wrong. To count the elements of a set with the ordinal omega you need > > > only the finite ordinals. > > > > I do not understand you. We cannot count all naturals without saying > > "omega" and being finished. > > I do not understand you. You cannot count all naturals and at some time > saying that you are finished. > (1) The set of natural numbers exist, by the axiom of infinity. > (2) You will never terminate when counting through the natural numbers. If we could not get to omega, we would not need it. And one would not be able to count omega + 1 and so on. Cantor could have refrained from introducing it. In order to count all finite sets you need omega. That is the essence of Cantor's theory. > > > > > Then we cannot say anything about f(oo). That is in fact the truth, but > > not in set theory. Set theory says about oo: omega is finished. > > Yes to the first, no to the second. f(oo) does not belong to set theory, > it belongs to analysis. Also note that Cantor at some time refrained from > using oo and started to use w, because there was too much confusion. > The oo from pre-Cantorian mathematics is different from the omega of > Cantorian mathematics. According to Cantor oo denotes potential infinity, omega denotes actual infinity. If all natural numbers exist, then the infinite oo of analysis in our vase problem is exactly the omega of set theory. > > === > > > > Any list gives a constructrible diagonal number. Any diagonal number is > > > > constructed: > > > > > > Yes. You can construct a number from any list, but you can not construct > > > a constructable number from a list that is itself not constructable. > > > > Every diagonal number is constructed and, therefore, is constructible. > > When constructed from a constructable list. What is an unconstructable list? Do you call any undefinite mess a list? Any list is construtced. Any diagonal number is constructed. > > > > > > > Fraenkel, Abraham A., Levy, Azriel: "Abstract Set Theory" (1976), p. > > > > 54: "Why, then, the restriction to the digits 1 and 2 in our proof? > > > > Just to kill the prejudice, found in some treatments of the proof, as > > > > if the method were purely existential, i.e. as if the proof, while > > > > showing that there exist decimals belonging to C but not to C0, did not > > > > allow to construct such decimals." > > > > > > What are C and C0 in this context? > > > > p. 52 explains it: "Lemma. Given any denumerable subset C0 of C, there > > exist members of C which are not contained in C0; that is to say, C0 is > > a proper subset of C." So C is the continuum R and C0 is the set of > > list numbers > > Yes, any diagonal number is constructed from the list. That does *not* make > any diagonal number constructible. Why then should it be important to stress that it is constructed? However, the set of constructed numbers is countable. > Obviously if the list is not > constructable, the diagonal number is constructed from the list, but is > itself not constructable. You apparently understand 'constructable' as > being able to be constructed from something, whatever that is. But > that term has a specific mathematical meaning. And constructible in the > context of numbers has a very specific meaning. > I understand: A number is constructible if it can be constructed. > > > > If you cannot understand that, then try this: The set of diagonal > > > > numbers is countable. Therefore the diagonal proof proves the > > > > uncountability of a countable set. > > > > > > Why is the set of diagonal numbers countable? How do you derive that? > > > > Because only a countable set of numbers can be constructed using a > > language with a finite alphabet. > > Using strings with a specific upperbound. > -- No. "Using finite strings" is enough. Regards, WM
From: mueckenh on 20 Oct 2006 17:03 Dik T. Winter schrieb: > In article <1161276574.792436.186750(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > In article <1161183237.727249.154740(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > > > > In article <1161079802.120515.175530(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > ... > > > > > > The inconsistency is that > > > > > > 1) For the balls inserted until noon, you can find the result: > > > > > > It is the set N. > > > > > > 2) For the balls removed until noon, you can find the result: > > > > > > It is the set N. > > > > > > 3) For the balls remaining at noon, the same arguments of > > > > > > continuity which lead to (1) and (2) cannot apply. > > > > > > > > > > There are quite a few obvious reasons. > > > > > (1) 1) is not because of continuity > > > > > > The reason that continuity plays no role is because the function of the > > > number of balls in the vase when written as a function of t is > > > discontinuous at infinitely many positions. > > > > But it is stepwise continuous. > > Yes, but it is infinitely many times not continuous in each neighbourhood > of the limit point. Let t be the ordinal number of transactions. It *is* continuous like a staircase. X(t) = 9 for t = 1 until t = 2 where X(t) switches to 18. That is enough to excflude X(omega) = 0. > > > > You can *model* the problem in different ways in mathematics, and one of > > > those models leads to 0 balls in the vase at time 0. And there are > > > presumably other models that give different results. So, unless the > > > problem is stated in a mathematically proper way, actually nothing can > > > be said about it. > > > > Translate balls as numbers and vase as set variable. More is not > > required. > > That is still not a mathematical formulation. More is required. You > need to actually state what you mean with 'number of balls in the vase > at noon' or 'natural numbers in the set at noon'. The number of transactions t is then t = omega at noon. Also there is a > time dependency that is not clearly stated. Time dependency can be eliminated if you count the transactions by natural numbers t as I do. > > If the vase is empty at noon, then can obtain from set theory that the > > limit of a sequence in no way can be determined by the terms of he > > sequence. > > The limit can in mathematics *always* be determined by the terms (if there > is a limit). But the limit in no case defines the function value at the > limit point. Then the irrational numbers as limit points are undefined. > > > Then there are no irrational numbers and several other parts > > of mathematics. > > The irrational numbers are by definition the limits of particular > sequencens. And that is not set theory. Everything is a set. Will you object that irrational numbers are defined on the basis of set theory? Regards, WM
From: MoeBlee on 20 Oct 2006 17:03
mueckenh(a)rz.fh-augsburg.de wrote: > "Continue in this manner" is just what is used to describe how the > infinite is realized. Not in formal axiomatic set theory. If you can't define your "continue in this manner" in the language of set theory, then your argument is not, contrary to your claim, in set theory, and so, if your argument does depend on "continue in this manner" and you won't define it in some set theory such as Z set theory, then, of course, we're done as far as me taking you seriously in regards to your argument and claim that it shows anything about set theory. MoeBlee |